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4 years ago

PLEASE HELP!!

1 answer:

4 0

Potential energy is stored energy due to your position. If comparing yourself to the ground (aka bottom of the hill), the higher you are, the higher the potential energy. Kinetic energy is energy due to motion.

That means at the top of the hill, you have the most potential energy and little/no kinetic energy, depending on if you started moving yet. If you're still at the top of the hill and not moving, you have zero kinetic energy. As you go down the hill, your potential energy decreases as it gets converted to kinetic energy. Once you're at the bottom of the hill, you should have the most kinetic energy and no potential energy because most of that energy was converted to kinetic energy.

Friction would decrease the amount of potential energy converted to kinetic energy because some of that energy would have been lost to friction, which moves in the opposite direction of your direction of motion. That means your speed at the bottom of the hill would be slower than if there was no friction present. That's the reason why you don't go flying off a slide when you get to the bottom, for example!

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Help me. It's question 2 I need the answer for

Alex73 [517]

Ok, let me see if I can help

Sound is caused by vibrations. These can pass through a solid, liquid, and gas. But not through vacuum because there are no particles

6 0

3 years ago

Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle

tangare [24]

Answer:

$F=1.159$

Explanation:

From the question we are told that:

Mass of pulley $M=1kg$

Radius $r=12cm$

Mass of block A $M_a=2.1kg$

Mass of block B $m_b=4.1kg$

Spring constant $\mu= 358 J/m2$

Generally the equation for Torque is mathematically given by

Since $\sumF=ma$

At mass A

$T_2-f_3=2.1a$

At mass B

$4.8-T_1=4.1a$

At Pulley

$R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}$

$R(T_1-T_2)=0.55a$

Therefore the equation for total force F

At mass A+At mass B+At Pulley

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a$

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a$

$-f_3+4.1=6.75a$

$-f_3=6.75a+4.8$

Since From above equation

$M_{eff}=6.7kg$

Therefore

$T=2\pi \sqrt{{\frac{M_{eff}}{k}}$

$T=2\pi \sqrt{{\frac{6.75}{\mu}}$

$T=0.862s$

Generally the equation for frequency is mathematically given by

$F=\frac{1}{T} \\F=\frac{1}{0.862}$

$F=1.159$

3 0

3 years ago

Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.

Nina [5.8K]

Not sure I’m really sorry

6 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

All waves on the electromagnetic spectrum travel at the same speed in a vacuum. The speed of these waves in a vacuum is 3x10 8 m

Alexandra [31]

As you said, the speed is constant in a vacuum, so the speed does not change, The frequency, on the other hand, decreases by half. The wavelength, λ (lambda), the frequency, v, and the speed of light, c, are related by the equation λν = c. We can see that frequency and wavelength are inversely proportional, so an increase in wavelength has a corresponding decrease in frequency.

6 0

3 years ago